cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A242174 Least prime divisor of A005260(n) which does not divide any previous term A005260(k) with k < n, or 1 if such a primitive prime divisor of A005260(n) does not exist.

Original entry on oeis.org

2, 3, 41, 5, 7, 349, 61, 75617, 31, 13, 499, 643897693, 17, 19, 1729774061, 101, 2859112064587, 138407, 83, 167, 59, 29, 653, 257, 997540809461453561581, 347, 13679, 37, 160449179727717672892660463, 211, 151, 43, 97, 73, 47
Offset: 1

Views

Author

Zhi-Wei Sun, May 07 2014

Keywords

Comments

Conjecture: a(n) is prime for any n > 0. In general, for any r > 2, if n is large enough then f_r(n) = sum_{k=0..n}C(n,k)^r has a prime divisor which does not divide any previous terms f_r(k) with k < n.

Examples

			a(3) = 41 since A005260(3) = 2^2*41 with 41 dividing none of A005260(1) = 2 and A005260(2) = 2*3^2.
		

Crossrefs

Programs

  • Mathematica
    u[n_]:=Sum[Binomial[n,k]^4,{k,0,n}]
    f[n_]:=FactorInteger[u[n]]
    p[n_]:=Table[Part[Part[f[n],k],1],{k,1,Length[f[n]]}]
    Do[If[u[n]<2,Goto[cc]];Do[Do[If[Mod[u[i],Part[p[n],k]]==0,Goto[aa]],{i,1,n-1}];Print[n," ",Part[p[n],k]];Goto[bb];Label[aa];Continue,{k,1,Length[p[n]]}];Label[cc];Print[n," ",1];Label[bb];Continue,{n,1,35}]

A242223 Least prime p such that H(n) == 0 (mod p) but H(k) == 0 (mod p) for no 0 < k < n, or 1 if such a prime p does not exist, where H(n) denotes the n-th harmonic number sum_{k=1..n}1/k.

Original entry on oeis.org

1, 3, 11, 5, 137, 7, 1, 761, 7129, 61, 97, 13, 29, 1049, 41233, 17, 37, 19, 7440427, 11167027, 18858053, 23, 583859, 577, 109, 34395742267, 521, 375035183, 4990290163, 31, 2667653736673, 2917, 269, 3583, 397, 1297, 10839223, 199, 737281, 41
Offset: 1

Views

Author

Zhi-Wei Sun, May 08 2014

Keywords

Comments

Conjecture: a(n) is prime except for n = 1, 7.

Examples

			a(4) = 5 since H(4) = 25/12 == 0 (mod 5), but none of H(1) = 1, H(2) = 3/2 and H(3) = 11/6 is congruent to 0 modulo 5.
		

Crossrefs

Programs

  • Mathematica
    h[n_]:=Numerator[HarmonicNumber[n]]
    f[n_]:=FactorInteger[h[n]]
    p[n_]:=Table[Part[Part[f[n],k],1],{k,1,Length[f[n]]}]
    Do[If[h[n]<2,Goto[cc]];Do[Do[If[Mod[h[i],Part[p[n],k]]==0,Goto[aa]],{i,1,n-1}];Print[n," ",Part[p[n],k]];Goto[bb];Label[aa];Continue,{k,1,Length[p[n]]}];Label[cc];Print[n," ",1];Label[bb];Continue,{n,1,40}]

A242292 Least prime divisor of 2^n - n which does not divide any 2^k - k with 0 < k < n, or 1 if such a primitive prime divisor of 2^n - n does not exist.

Original entry on oeis.org

1, 2, 5, 3, 1, 29, 11, 31, 503, 13, 7, 1021, 8179, 1637, 4679, 1, 8737, 131063, 524269, 262139, 2097131, 349, 131, 773, 271, 197, 457, 1493, 317, 17, 6733, 73, 41, 157109, 83, 53, 1741, 3329, 49977801259, 997, 149, 2199023255531, 61, 4398046511093, 3769453
Offset: 1

Views

Author

Zhi-Wei Sun, May 10 2014

Keywords

Comments

Conjecture: a(n) = 1 only for n = 1, 5, 16.
In constrast, a classical theorem of Bang asserts that if n > 1 is different from 6 then 2^n - 1 has a prime divisor which does not divide any 2^k - 1 with 0 < k < n.

Examples

			a(4) = 3 since 2^4 - 4 = 2^2*3 with 3 dividing none of 2^1 - 1 = 1, 2^2 - 2 = 2 and 2^3 - 3 = 5.
		

References

  • A. S. Bang, Taltheoretiske Undersgelser, Tidsskrift fur Mat. 4(1886), no. 5, 70--80, 130--137.

Crossrefs

Programs

  • Mathematica
    u[n_]:=2^n-n
    f[n_]:=FactorInteger[u[n]]
    p[n_]:=Table[Part[Part[f[n], k], 1], {k, 1, Length[f[n]]}]
    Do[If[u[n]<2, Goto[cc]]; Do[Do[If[Mod[u[i], Part[p[n], k]]==0, Goto[aa]], {i, 1, n-1}]; Print[n, " ", Part[p[n], k]]; Goto[bb]; Label[aa]; Continue, {k, 1, Length[p[n]]}]; Label[cc]; Print[n, " ", 1]; Label[bb]; Continue, {n, 1, 45}]
Showing 1-3 of 3 results.